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Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

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10 Diagnostic Tests 630 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #123 : Solving Quadratic Equations

Solve the quadratic equation by completing the square:

x^2+22x+100=0

Possible Answers:

$x=-9.01\text{ or }-18.23$

$x=-9.29\text{ or }-21.32$

$x=-5.45\text{ or }-15.61$

$x=-6.42\text{ or }-15.58$

Correct answer:

$x=-6.42\text{ or }-15.58$

Explanation:

Start by moving the number to the right of the equation so that all the terms with values are alone:

x^2+22x+100=0

x^2+22x=-100

Now, you need to figure out what number to add to both sides. To do so, take the coefficient in front of the term, divide it by , then square it:

Coefficient:

$\text{Term to add}=(\frac{22}{2})^2=(11)^2=121$

Add this number to both sides of the equation:

x^2+22x+121=-100+121=21

Simplify the left side of the equation.

(x+11)^2=21

Now solve for .

$x+11=\pm\sqrt{21}$

$x=-11+\sqrt{21}=-6.42$ , or

$x=-11-\sqrt{21}=-15.58$

Make sure to round to places after the decimal.

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Example Question #55 : Completing The Square

Solve the quadratic equation by completing the square:

x^2-30x+98=0

Possible Answers:

$x=21.92\text{ or }12.28$

$x=26.27\text{ or }3.73$

$x=29.71\text{ or }9.22$

$x=25.67\text{ or }8.21$

Correct answer:

$x=26.27\text{ or }3.73$

Explanation:

Start by moving the number to the right of the equation so that all the terms with values are alone:

x^2-30x+98=0

x^2-30x=-98

Now, you need to figure out what number to add to both sides. To do so, take the coefficient in front of the term, divide it by , then square it:

Coefficient:

$\text{Term to add}=(\frac{30}{2})^2=(15)^2=225$

Add this number to both sides of the equation:

x^2-30x+225=-98+225=127

Simplify the left side of the equation.

(x-15)^2=127

Now solve for .

$x-15=\pm\sqrt{127}$

$x=15+\sqrt{127}=26.27$ , or

$x=15-\sqrt{127}=3.73$

Make sure to round to places after the decimal.

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Example Question #56 : Completing The Square

Solve the following quadratic equation by completing the square:

x^2-16x+40=0

Possible Answers:

$x=-5.19\text{ or }10.72$

$x=4.19\text{ or }19.52$

$x=-3.09\text{ or }14.98$

$x=3.10\text{ or }12.90$

Correct answer:

$x=3.10\text{ or }12.90$

Explanation:

Start by moving the number to the right of the equation so that all the terms with values are alone:

x^2-16x+40=0

x^2-16x=-40

Now, you need to figure out what number to add to both sides. To do so, take the coefficient in front of the term, divide it by , then square it:

Coefficient:

$\text{Term to add}=(\frac{-16}{2})^2=(-8)^2=64$

Add this number to both sides of the equation:

x^2-16x+64=-40+64=24

Simplify the left side of the equation.

(x-8)^2=24

Now solve for .

$x-8=\pm\sqrt{24}$

$x=8+\sqrt{24}=12.90$ , or

$x=8-\sqrt{24}=3.10$

Make sure to round to places after the decimal.

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Example Question #1581 : Algebra Ii

Which of the following is the same after completing the square?

3x^2+x-6 =0

Possible Answers:

$(x+\frac{1}{6})^2 =\frac{73}{36}$

$(x+\frac{1}{3})^2 =\frac{7}{2}$

$(x-\frac{1}{6})^2 =\frac{73}{36}$

$(x-\frac{1}{3})^2 =\frac{7}{2}$

$(x+\frac{1}{6})^2 =\frac{1}{2}$

Correct answer:

$(x+\frac{1}{6})^2 =\frac{73}{36}$

Explanation:

Divide by three on both sides.

$\frac{3x^2+x-6 }{3}=\frac{0}{3}$

$x^2+\frac{1}{3}x-2 = 0$

Add two on both sides.

$x^2+\frac{1}{3}x-2 +2= 0+2$

$x^2+\frac{1}{3}x = 2$

To complete the square, we will need to divide the one-third coefficient by two, which is similar to multiplying by one half, square the quantity, and add the two values on both sides.

$x^2+\frac{1}{3}x+ (\frac{1}{3}\cdot \frac{1}{2})^2 = 2+ (\frac{1}{3}\cdot \frac{1}{2})^2$

Simplify both sides.

$x^2+\frac{1}{3}x+ \frac{1}{36} = 2+ \frac{1}{36}$

Factor the left side, and combine the terms on the right.

$(x+\frac{1}{6})^2 = \frac{72}{36}+ \frac{1}{36}$

The answer is: $(x+\frac{1}{6})^2 =\frac{73}{36}$

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Example Question #1581 : Algebra Ii

Solve for by completing the square.

x^2+8x-2=0

Possible Answers:

$x=0.39\text{ and }x=-4.39$

$x=1.24\text{ and }x=-10.24$

$x=0.24\text{ and }x=-8.24$

$x=0.97\text{ and }x=-8.97$

Correct answer:

$x=0.24\text{ and }x=-8.24$

Explanation:

x^2+8x-2=0

Start by adding to both sides so that the terms with the are together on the left side of the equation.

x^2+8x=2

Now, look at the coefficient of the -term. To complete the square, divide this coefficient by , then square the result. Add this term to both sides of the equation.

x^2+8x+16=2+16

Rewrite the left side of the equation in the squared form.

(x+4)^2=18

Take the square root of both sides.

$x+4=\pm \sqrt{18}$

Now solve for .

$x=-4+\sqrt{18}=0.24$

$x=-4-\sqrt{18}=-8.24$

Round to two places after the decimal.

Report an Error

Example Question #1582 : Algebra Ii

Solve for by completing the square.

x^2-6x+2=0

Possible Answers:

$x=5.65\text{ and }x=0.35$

$x=1.95\text{ and }x=5.05$

$x=-8.75\text{ and }x=-2.25$

$x=10.55\text{ and }x=2.45$

Correct answer:

$x=5.65\text{ and }x=0.35$

Explanation:

x^2-6x+2=0

Start by subtracting from both sides so that the terms with the are together on the left side of the equation.

x^2-6x=-2

Now, look at the coefficient of the -term. To complete the square, divide this coefficient by , then square the result. Add this term to both sides of the equation.

x^2-6x+9=-2+9

Rewrite the left side of the equation in the squared form.

(x-3)^2=7

Take the square root of both sides.

$x-3=\pm \sqrt{7}$

Now solve for .

$x=3+\sqrt{7}=5.65$

$x=3-\sqrt{7}=0.35$

Round to two places after the decimal.

Report an Error

Example Question #1582 : Algebra Ii

Solve for by completing the square.

x^2-10x+20=0

Possible Answers:

$x=3.33\text{ and }x=8.67$

$x=-5.21\text{ and }x=-9.79$

$x=8.06\text{ and }x=1.94$

$x=7.24\text{ and }x=2.76$

Correct answer:

$x=7.24\text{ and }x=2.76$

Explanation:

x^2-10x+20=0

Start by subtracting from both sides so that the terms with the are together on the left side of the equation.

x^2-10x=-20

Now, look at the coefficient of the -term. To complete the square, divide this coefficient by , then square the result. Add this term to both sides of the equation.

x^2-10x+25=-20+25

Rewrite the left side of the equation in the squared form.

(x-5)^2=5

Take the square root of both sides.

$x-5=\pm \sqrt{5}$

Now solve for .

$x=5+\sqrt{5}=7.24$

$x=5-\sqrt{5}=2.76$

Round to two places after the decimal.

Report an Error

Example Question #442 : Intermediate Single Variable Algebra

Solve for by completing the square.

2x^2-8x+1=0

Possible Answers:

$x=2.91\text{ and }x=0.09$

$x=8.57\text{ and }x=6.43$

$x=1.37\text{ and }x=5.63$

$x=3.87\text{ and }x=0.13$

Correct answer:

$x=3.87\text{ and }x=0.13$

Explanation:

2x^2-8x+1=0

Start by subtracting from both sides so that the terms with the are together on the left side of the equation.

2x^2-8x=-1

Next, divide everything by the coefficient of the x^2 term.

$x^2-4x=-\frac{1}{2}$

Now, look at the coefficient of the -term. To complete the square, divide this coefficient by , then square the result. Add this term to both sides of the equation.

$x^2-4x+4=-\frac{1}{2}+4$

Rewrite the left side of the equation in the squared form.

$(x-2)^2=\frac{7}{2}$

Take the square root of both sides.

$x-2=\pm \sqrt{\frac{7}{2}}$

Now solve for .

$x=2+\sqrt{\frac{7}{2}}=3.87$

$x=5-\sqrt{\frac{7}{2}}=0.13$

Round to two places after the decimal.

Report an Error

Example Question #281 : Quadratic Equations And Inequalities

Solve for by completing the square.

3x^2-5x-10=0

Possible Answers:

$x=3.69\text{ and }x=-2.31$

$x=7.08\text{ and }x=-3.92$

$x=4.09\text{ and }x=1.91$

$x=2.84\text{ and }x=-1.17$

Correct answer:

$x=2.84\text{ and }x=-1.17$

Explanation:

3x^2-5x-10=0

Start by adding to both sides so that the terms with the are together on the left side of the equation.

3x^2-5x=10

Next, divide everything by the coefficient of the x^2 term.

$x^2-\frac{5}{3}x=\frac{10}{3}$

Now, look at the coefficient of the -term. To complete the square, divide this coefficient by , then square the result. Add this term to both sides of the equation.

$x^2-\frac{5}{3}x+\frac{25}{36}=\frac{10}{3}+\frac{25}{36}$

Rewrite the left side of the equation in the squared form.

$(x-\frac{5}{6})^2=\frac{145}{36}$

Take the square root of both sides.

$x-\frac{5}{6}=\pm \sqrt{\frac{145}{36}}$

Now solve for .

$x=\frac{5}{6}+\sqrt{\frac{145}{36}}=2.84$

$x=\frac{5}{6}-\sqrt{\frac{145}{36}}=-1.17$

Round to two places after the decimal.

Report an Error

Example Question #63 : Completing The Square

Solve for by completing the square.

x^2-10x-12=0

Possible Answers:

$x=20.75\text{ and }x=8.75$

$x=12.36\text{ and }x=2.36$

$x=11.08\text{ and }x=-1.08$

$x=-3.46\text{ and }x=-16.46$

Correct answer:

$x=11.08\text{ and }x=-1.08$

Explanation:

x^2-10x-12=0

Start by adding to both sides so that the terms with the are together on the left side of the equation.

x^2-10x=12

Now, look at the coefficient of the -term. To complete the square, divide this coefficient by , then square the result. Add this term to both sides of the equation.

x^2-10x+25=12+25

Rewrite the left side of the equation in the squared form.

(x-5)^2=37

Take the square root of both sides.

$x-5=\pm \sqrt{37}$

Now solve for .

$x=5+\sqrt{37}=11.08$

$x=5-\sqrt{37}=-1.08$

Round to two places after the decimal.

Report an Error

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Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #123 : Solving Quadratic Equations

Example Question #55 : Completing The Square

Example Question #56 : Completing The Square

Example Question #1581 : Algebra Ii

Example Question #1581 : Algebra Ii

Example Question #1582 : Algebra Ii

Example Question #1582 : Algebra Ii

Example Question #442 : Intermediate Single Variable Algebra

Example Question #281 : Quadratic Equations And Inequalities

Example Question #63 : Completing The Square

All Algebra II Resources

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